Physics Homework Problem Newtons Laws of Motion

I have this problem on a web assignment due on the 20th of this month. A few of the problems have completely blown my mind including the following:

A force F0 causes an acceleration of 18 m/s2 when it acts on an object of mass m sliding on a frictionless surface. Find the magnitude of the acceleration of the same object in the circumstances shown in each figure below

I know F=ma, but doesn't seem to be enough to solve this problem. I have also included a picture of the diagram which accompanied this problem. I would appreciate any help or guidance that could point me in the right direction.

Do this by drawing a diagram in which the two forces form two sides of a triangle. The third side represents the resultant force.

Since you know two of the side lengths, and also the angle which separates them, it is possible to calculate the magnitude of the third force using the cosine rule.
[tex]c^{2} = a^{2} + b^{2} - 2abCosC[/tex]

It is also possible to determine the angle that the resultant force acts at by using the sine rule.
a/SinA = b/SinB = c/SinC

If I think that I understand you correctly ur saying to use the law of cosines to find the magnitude of the resultant force graphically. That makes sense, but what I don't understand is how that would work. Only the acceleration is given, the magnitude of the force is not given in this problem. Also, the mass of the box is not given so how would the law of cosines be applicable here?

never mind...makes sense now. I feel like a total retard. I used Pythagorean theorum to solve part A. I just realized that the pythagorean theorum is becomes a special case of the law of cosines when angle C = 90*

Well for the first case you have all you need, you should be able to intuitively "see" the resultant force. Now take the component of each applied force in that direction, using this and the known force [tex]F_0[/tex] you can solve for the unknown acceleration.

Once you have the acceleration you can solve for mass m and use it in part b.